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# Smart numbers

Author Message
CEO
Joined: 21 Jan 2007
Posts: 2734

Kudos [?]: 1076 [0], given: 4

Location: New York City

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29 Oct 2007, 13:04
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How can I solve this using smart numbers

Each of the N members of the club agreed to contribute X dollars to buy a new computer. IF the club contained 2 more members, HOW MUCH LESS would everyone have to contribute?

NX / (N+2)
X [(N+N) / (N+2)]
N / (XN + 2X)
X [(1-N) / (N+2)]
XN - [N/(N+2)]

Kudos [?]: 1076 [0], given: 4

SVP
Joined: 29 Aug 2007
Posts: 2471

Kudos [?]: 857 [0], given: 19

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29 Oct 2007, 15:07
bmwhype2 wrote:
How can I solve this using smart numbers

Each of the N members of the club agreed to contribute X dollars to buy a new computer. IF the club contained 2 more members, HOW MUCH LESS would everyone have to contribute?

NX / (N+2)
X [(N+N) / (N+2)]
N / (XN + 2X)
X [(1-N) / (N+2)]
XN - [N/(N+2)]

before:
number of people = n
each person's contribution = x
total contribution = nx

after:
number of people = n+2
each person's contribution = nx/(n+2)
total contribution = nx

so the difference = x - nx/(n+2) = x [1 - (n)/(n+2)]

so none of the above. D seems close if there is any typo.

Kudos [?]: 857 [0], given: 19

CEO
Joined: 21 Jan 2007
Posts: 2734

Kudos [?]: 1076 [0], given: 4

Location: New York City

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20 Dec 2007, 12:12
GMAT TIGER wrote:
bmwhype2 wrote:
How can I solve this using smart numbers

Each of the N members of the club agreed to contribute X dollars to buy a new computer. IF the club contained 2 more members, HOW MUCH LESS would everyone have to contribute?

NX / (N+2)
X [(N+N) / (N+2)]
N / (XN + 2X)
X [(1-N) / (N+2)]
XN - [N/(N+2)]

before:
number of people = n
each person's contribution = x
total contribution = nx

after:
number of people = n+2
each person's contribution = nx/(n+2)
total contribution = nx

so the difference = x - nx/(n+2) = x [1 - (n)/(n+2)]

so none of the above. D seems close if there is any typo.

sorry you are correct, it is x [1 - (n)/(n+2)]
i need to be more careful interpreting the algebra

Kudos [?]: 1076 [0], given: 4

Re: Smart numbers   [#permalink] 20 Dec 2007, 12:12
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# Smart numbers

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